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3 years ago

As a cold air mass advances on a warm air mass, what usually comes before it?

1 answer:

8 0

Answer: A cold front occurs when a cold air mass advances into a region occupied by a warm air mass. If the boundary between the cold and warm air masses doesn't move, it is called a stationary front.

Explanation: Two types of occluded front exist: the warm-type and the cold-type. They’re distinguished by the relative temperatures of the air mass ahead of the occlusion – in other words, the air mass ahead of the original warm front – and the air mass behind the cold front. If the air behind the cold front is colder than the air ahead of the occlusion, it shoves beneath that air (because it’s denser) to form a cold-type occluded front. If the air behind the cold front is warmer than the air ahead, it rides over it to form a warm-type occluded front – which appears to be the more common case. In either situation, the lighter warm air representing the air mass originally between the warm and cold fronts sits above the boundary between the two cooler air masses.

Hope this helps!!

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On earth, two parts of a space probe weigh 14500 N and 4800 N. These parts are separated by a center-to-center distance of 18 m

Nastasia [14]

Answer:

$F = 1.489*10^{-7} N$

Explanation: Weight of space probes on earth is given by: $W= m*g$

W= weight of the object( in N)

m= mass of the object (in kg)

g=acceleration due to gravity(9.81 $\frac{m}{s^{2} }$ )

Therefore,

$m_{1} = \frac{14500}{9.81}$

$m_{1} = 1478.08 kg$

Similarly,

$m_{2} = \frac{4800}{9.81}$

$m_{2} = 489.29 kg$

Now, considering these two parts as uniform spherical objects

Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.

Force between these two objects is given by:

$F = \frac{Gm_{1} m_{2}}{R^{2} }$

G= gravitational constant ( $6.67 * 10^{-11} m^{3} kg^{-1} s^{-2}$ )

$m_{1} , m_{2}$ = masses of the object

R= distance between their centres (in m)(18 m)

Substituiting all these values into the above formula

$F = 1.489*10^{-7} N$

This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.

7 0

3 years ago

You walk to school every day. It is 0.5 miles. After school you walk to Kroger, which is 1.3 miles. You leave Kroger to go all b

marin [14]

The total distance you haved walked should be 2.6 miles

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2 years ago

1) A train accelerates from 36 km/hr to 54 km/hr in 10<br> s. Find acceleration?

Crank

Answer: Given:

Initial velocity= 36km/h=36x5/18=10m/s

Final velocity =54km/h=54x5/18=15m/s

Time =10sec

Acceleration = v-u/ t

=15-10/10=5/10=1/2=0.5 m/s2

Distance =s=?

From second equation of motion:

S=ut +1/2 at^2

=10*10+1/2*0.5*10*10

=100+25

=125m

So distance travelled 125m

Hope it helps you

3 0

3 years ago

A truck moving at 13.3 m/s hits a concrete wall. As a result of the collision, a 6-kg wrench moves forwards and strikes the wall

Naddik [55]

Answer:

Explanation:

The velocity of the wrench must be equal to the velocity of the truck . So momentum of the wrench before it hits the wall

= mv = 6 x 13.3 = 79.8 kg m /s

If resisting force of wall be F , impulse on the wrench = F x time

= F x .07

Impulse = change in momentum of the wrench = mv - 0 = mv = 79.8 kgm/s

So F x .07 = 79.8

F = 1140 N .

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3 years ago

A person with a mass of 40 kg is sitting on a box. What is the value of the normal force

VMariaS [17]

normal force=mass*gravitational force

normal force=40*0

normal force=40

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2 years ago

As a cold air mass advances on a warm air mass, what usually comes before it?​

As a cold air mass advances on a warm air mass, what usually comes before it?